Meter Stick Challenge

My initial plan to solve this challenge was very vague. I was not sure how to tackle the problem at all but I did know that it had something to do with torque. Since I knew the equation for torque was lever arm multiplied by force that is the first thing that I wrote down. Since I knew that there was two torques in this situation I wrote lever arm times force = lever arm times force. I knew that the weight was one hundred grams and since that is mass instead of force I calculated that the force of the weight was .98 newtons. From there I was stuck. However, my group members helped me figure out that the center of gravity of the stick was at the 50 centimeter mark because that is the point at which it balanced like a sea saw on the edge of the table. With the weight on the end, the center of mass was at the 25 centimeter mark. Therefore, we figured that the lever arm lengths were 25 and 75 centimeters. However, we realized that the center of gravity is where the counter clockwise lever arm began. From there, we filled in the equation and the only variable we had left was the force for the clock wise torque. We solved for that and converted it from newtons to grams and got that the meter stick was 100 grams. When we weighed it we found that it was 101 grams. The reason that the meter stick could balance on the center of mass and the center of gravity because in both cases there was an equal force on each side.